我有JavaScript，它周期性地执行活动。当用户没有在看站点时(例如，窗口或选项卡没有焦点)，最好不要运行。
是否有一种方法可以使用JavaScript来做到这一点?
我的参考点:如果你正在使用的窗口不活跃，Gmail聊天就会播放声音。
当前回答
简单的/立即检查:
if(document.hidden) {
// do something
}
可见性变化事件:
document.addEventListener("visibilitychange", function() {
console.log(document.visibilityState); // "hidden" or "visible"
}, false);
基于承诺的事件:
// An `await`able function that resolves when page visibility changes:
function visibilityChange(state="") {
return new Promise(resolve => {
document.addEventListener("visibilitychange", function() {
if(!state || document.visibilityState === state) {
resolve(document.visibilityState);
document.removeEventListener("visibilitychange", arguments.callee);
}
});
});
}
// Use it like this:
await visibilityChange();
console.log(document.visibilityState);
// Or wait for page to become...
await visibilityChange("visible");
await visibilityChange("hidden");
(注:后两个解是我加到这个答案里的，因为这个问题现在已经结束了，我不能加我自己的答案。以防有人认为我抄袭了那篇文章而没有注明出处。)
其他回答
我为我的应用程序创建了一个Comet Chat，当我收到来自我使用的另一个用户的消息时:
if(new_message){
if(!document.hasFocus()){
audio.play();
document.title="Have new messages";
}
else{
audio.stop();
document.title="Application Name";
}
}
var visibilityChange = (function (window) {
var inView = false;
return function (fn) {
window.onfocus = window.onblur = window.onpageshow = window.onpagehide = function (e) {
if ({focus:1, pageshow:1}[e.type]) {
if (inView) return;
fn("visible");
inView = true;
} else if (inView) {
fn("hidden");
inView = false;
}
};
};
}(this));
visibilityChange(function (state) {
console.log(state);
});
http://jsfiddle.net/ARTsinn/JTxQY/
简单的/立即检查:
if(document.hidden) {
// do something
}
可见性变化事件:
document.addEventListener("visibilitychange", function() {
console.log(document.visibilityState); // "hidden" or "visible"
}, false);
基于承诺的事件:
// An `await`able function that resolves when page visibility changes:
function visibilityChange(state="") {
return new Promise(resolve => {
document.addEventListener("visibilitychange", function() {
if(!state || document.visibilityState === state) {
resolve(document.visibilityState);
document.removeEventListener("visibilitychange", arguments.callee);
}
});
});
}
// Use it like this:
await visibilityChange();
console.log(document.visibilityState);
// Or wait for page to become...
await visibilityChange("visible");
await visibilityChange("hidden");
(注:后两个解是我加到这个答案里的，因为这个问题现在已经结束了，我不能加我自己的答案。以防有人认为我抄袭了那篇文章而没有注明出处。)
我重读了@daniel-buckmaster的版本 我没有做多次尝试，但是，代码对我来说似乎更优雅…
// on-visibility-change.js v1.0.1, based on https://stackoverflow.com/questions/1060008/is-there-a-way-to-detect-if-a-browser-window-is-not-currently-active#38710376
function onVisibilityChange(callback) {
let d = document;
let visible = true;
let prefix;
if ('hidden' in d) {
prefix = 'h';
} else if ('webkitHidden' in d) {
prefix = 'webkitH';
} else if ('mozHidden' in d) {
prefix = 'mozH';
} else if ('msHidden' in d) {
prefix = 'msH';
} else if ('onfocusin' in d) { // ie 9 and lower
d.onfocusin = focused;
d.onfocusout = unfocused;
} else { // others
window.onpageshow = window.onfocus = focused;
window.onpagehide = window.onblur = unfocused;
};
if (prefix) {
visible = !d[prefix + 'idden'];
d.addEventListener(prefix.substring(0, prefix.length - 1) + 'visibilitychange', function() {
(d[prefix + 'idden'] ? unfocused : focused)();
});
};
function focused() {
if (!visible) {
callback(visible = true);
};
};
function unfocused() {
if (visible) {
callback(visible = false);
};
};
};
U可以使用:
(function () {
var requiredResolution = 10; // ms
var checkInterval = 1000; // ms
var tolerance = 20; // percent
var counter = 0;
var expected = checkInterval / requiredResolution;
//console.log('expected:', expected);
window.setInterval(function () {
counter++;
}, requiredResolution);
window.setInterval(function () {
var deviation = 100 * Math.abs(1 - counter / expected);
// console.log('is:', counter, '(off by', deviation , '%)');
if (deviation > tolerance) {
console.warn('Timer resolution not sufficient!');
}
counter = 0;
}, checkInterval);
})();
